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Circuit with DC & AC sources

  1. Jan 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Find thevenin's equivalent for the capacitor between node a and b and determine Vc (image attached)

    2. Relevant equations
    I know I can use superposition to get Voc, but when I add the two answers together, can I combine them into a single source or keep them seperated?

    3. The attempt at a solution
    After finding Thevenin's equivalent (E = 12V + 24V<0) & (Z = 9 - j1 ohms <~ after I plug the capacitor back in), I THINK i need to keep the sources seperate and do another superposition. so using the DC source: the capacitor acts as an open, meaning Vc= 12V, and using the AC source: a voltage divider: (24V<0)(-j1/(9 - j1)).

    answer: 12V + 2.64V<-83.7

    Attached Files:

    Last edited: Jan 14, 2007
  2. jcsd
  3. Jan 14, 2007 #2
    upload your image on imageshack or something. Sometimes it takes awhile to get an image approved if you upload it on the forum
  4. Jan 14, 2007 #3
    okay, i was unaware

  5. Jan 14, 2007 #4
    Yes, that should be it.
  6. Jan 15, 2007 #5


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    Not sure what you are actually asking but ....

    To get thevenin's equivalent just need to know two things: [tex]V_T, R_T[/tex].

    since there are no dependent sources, you can work out [tex]R_T[/tex] by simply zeroing all sources. voltage source->short circuit, current source->open circuit. and then look back from outside terminals a-b to see what is the resistance in there. that's your [tex]R_T[/tex]. I assumed your [tex]1\Omega[/tex] for the capacitor is actually [tex]-j\Omega[/tex] from your diagram. The phase you have written down for this seems correct but the magitute seems not (should check again).

    For the [tex]V_T[/tex], it is better to work out short-circuit current first, then [tex]V_T=i_{sc} R_T[/tex]. Use node voltage analysis. And by the way, it is not 12V! you have two sources there... unless it is one big coincidence.
  7. Jan 15, 2007 #6
    none of my answers were simply "12V". For the thevenin's voltage I got:

    12V + 24V<0

    which brings me back to my ORIGINAL question (this is what I was actually asking) about whether you can combine this DC and AC voltage into a single voltage source or keep them seperated.

    as for Rth, i got 9 ohms
    Last edited: Jan 15, 2007
  8. Jan 15, 2007 #7


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    ok, now I understand what you are asking. ignore my previous post. Now whether you wanna combine them, it is just how you may want to "draw" the diagram you can have one voltage with
    [tex]\displaymath{V_{T}(t)=12 + 24\cos (\omega t + \phi) V}[/tex]
    now here [tex]\phi[/tex] is 0 if that's what your phasor means or -90 if your convention is to use [tex]\sin[/tex] instead.

    alternatively, you can have one phasor source and one DC source. Don't think it is that important which one you use to represent it. but remember though you can't really "absorb" the constant 12V. think about it.... it is like shifting the entire sinusoid up.

    and no, I didn't get [tex]R_T[/tex] as real.
  9. Jan 16, 2007 #8
    care to explain?

    shorting the voltage source and opening the current source leaves a 6 ohm resistor and a 3 ohm resistor when looking at the opened capacitor? maybe we're trying to find different things, but 9 ohms is my answer for thevenin's resistance, and (9 ohms - j1 ohms) is the total resistance with the capacitor plugged back into thevenin's equivalent.

    That's one "you're right" and one "you're wrong" for my answer on this problem

    anyone want to break the tie?
    Last edited: Jan 16, 2007
  10. Jan 16, 2007 #9


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    Well the way I see it, [itex]R_{th}[/itex] is 9 ohms. And when you plug the capacitor back into the thevenin circuit, the total impedance, Z, is [itex]R_{th} + X_c[/itex] which gives 9 ohms - j1 ohms.
  11. Jan 16, 2007 #10


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    ok, we could have been looking at the circuit slightly differently, I included the capacitor in every steps, ie. assumming that the capacitor has to be integrated in the equivalent circuit as well. Since you have phasor sources, my understanding is that you can't treat capacitors as open circuit once you have zeroed the sources for the purpose of working out [tex]R_T[/tex]. anyway, if the capacitor is supposed to be excluded from the equivalent circuit then what you have done looks fine to me (I have not double checked your numerical answers though... but method is correct)
  12. Jan 16, 2007 #11


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    When you do thevenin for this circuit, the load (capacitor) has to be removed. There is no need to include it in steps to find Vth or Rth. Only after the equivalent circuit is found then you include it if you wish to find things such as total impedance.
  13. Jan 16, 2007 #12


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    ok, I didn't look at it that way , but what you said is correct. I thought the terminals a-b are the external terminals that is to be attached to some load....(not shown in diagram) and hence I interpreted the capacitor as part of circuit....and treated the problem as "given this circuit, turn it into a Thevenin equivalent about the terminals a,b"... But obviously, both of u saw the capacitor as a load instead, .... ok, may be the diagram should have been better drawn or someone should tell me that the capacitor is actually the load. Sorry for the confusion.
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